What is the extreme f(x,y)=3x^2-9x+9xy^2 ?

The extreme f(x,y)=3x^2-9x+9xy^2 is Saddle(0,1),Saddle(0,-1),Minimum(3/2 ,0)

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extreme f(x,y)=3x^2-9x+9xy^2

extreme points

f (x, y) = 3 x^{2} - 9 x + 9 x y^{2}

Saddle (0, 1), Saddle (0, - 1), Minimum (\frac{3}{2}, 0)

Solution steps

Find the critical points:

(0, 1), (0, - 1), (\frac{3}{2}, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 18 x

D (x, y) = 108 x - 324 y^{2}

Check the sign of at each critical point

Check critical point

(0, 1) :

Saddle

Check critical point

(0, - 1) :

Saddle

Check critical point

(\frac{3}{2}, 0) :

Minimum

Saddle (0, 1), Saddle (0, - 1), Minimum (\frac{3}{2}, 0)

2 θ - 4 sin (θ)

f (1.1, 1.1) = 2 x + 3 y - x^{3} - 2 y^{2}

f (x) = 9 - x

f (x) = x^{9} e^{2 x}

f (x) = x (18 - 40 + 2 x) (\frac{40}{2} - x)

What is the extreme f(x,y)=3x^2-9x+9xy^2 ?
The extreme f(x,y)=3x^2-9x+9xy^2 is Saddle(0,1),Saddle(0,-1),Minimum(3/2 ,0)